Plus One Physics Model Paper

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Kerala State Syllabus - Plus One Physics Model Question Paper - Answers Part A: Answer all questions. Each question carries 1 mark. Which of the following is a fundamental unit? (a) Newton (b) Joule (c) Kilogram (d) Watt Answer: (c) Kilogram State the principle of homogeneity of dimensions. Answer: The principle of homogeneity of dimensions states that an equation is dimensionally correct if the dimensions of the terms on both sides of the equation are the same. Only quantities of the same dimensions can be added or subtracted. A body is thrown vertically upwards. What is its velocity at the highest point? Answer: Zero What is the angle between $\vec{A}$ and $\vec{B}$ if $\vec{A} \cdot \vec{B} = 0$? Answer: $90^\circ$ (assuming $\vec{A}$ and $\vec{B}$ are non-zero vectors) State the work-energy theorem. Answer: The work-energy theorem states that the net work done by all forces acting on an object is equal to the change in the object's kinetic energy. ($W_{net} = \Delta K$) Give an example of a non-conservative force. Answer: Friction (or air resistance, viscous force) What is the moment of inertia of a circular disc about an axis passing through its center and perpendicular to its plane? Answer: $I = \frac{1}{2}MR^2$ State Hooke's Law. Answer: Hooke's Law states that within the elastic limit, the stress applied to a material is directly proportional to the strain produced in it. (Stress $\propto$ Strain) For a satellite orbiting close to the Earth's surface, what is its orbital velocity? Answer: $v_o = \sqrt{gR_e}$ What is the SI unit of coefficient of viscosity? Answer: Pascal-second (Pa s) or N s/m$^2$ Part B: Answer any 8 questions. Each question carries 2 marks. The percentage error in the measurement of mass and speed are $2\%$ and $3\%$ respectively. How much will be the maximum percentage error in the estimate of kinetic energy obtained by measuring mass and speed? Answer: Kinetic Energy $K = \frac{1}{2}mv^2$. Percentage error in $K$: $\frac{\Delta K}{K} \times 100 = \frac{\Delta m}{m} \times 100 + 2 \frac{\Delta v}{v} \times 100$ $= 2\% + 2(3\%) = 2\% + 6\% = 8\%$. Maximum percentage error is $8\%$. Derive the relation $v = u + at$ using calculus method. Answer: Acceleration $a = \frac{dv}{dt}$ $dv = a \ dt$ Integrating both sides: $\int_{u}^{v} dv = \int_{0}^{t} a \ dt$ $[v]_{u}^{v} = a [t]_{0}^{t}$ (assuming 'a' is constant) $v - u = at - 0$ $v = u + at$ A car moving with a speed of $36 \text{ km/h}$ is brought to rest in $10 \text{ m}$ by applying brakes. Calculate the retardation produced by the brakes. Answer: Given: $u = 36 \text{ km/h} = 36 \times \frac{5}{18} = 10 \text{ m/s}$, $v = 0 \text{ m/s}$, $s = 10 \text{ m}$. Using $v^2 = u^2 + 2as$: $0^2 = (10)^2 + 2a(10)$ $0 = 100 + 20a$ $20a = -100$ $a = -5 \text{ m/s}^2$. Retardation is $5 \text{ m/s}^2$. What is the difference between elastic collision and inelastic collision? Give one example for each. Answer: Elastic Collision: Both momentum and kinetic energy are conserved. Example: Collision between billiard balls. Inelastic Collision: Momentum is conserved, but kinetic energy is NOT conserved (some kinetic energy is lost as heat, sound, or deformation). Example: A bullet embedding itself in a block of wood. Define angular momentum. State the law of conservation of angular momentum. Answer: Angular Momentum ($L$): It is the rotational analogue of linear momentum. For a particle, it is defined as the cross product of its position vector ($\vec{r}$) and linear momentum ($\vec{p}$): $\vec{L} = \vec{r} \times \vec{p}$. For a rigid body, it is $L = I\omega$. Law of Conservation of Angular Momentum: If no external torque acts on a system, its total angular momentum remains conserved. ($I_1\omega_1 = I_2\omega_2$) Distinguish between streamline flow and turbulent flow. Answer: Streamline Flow (Laminar Flow): The fluid particles follow smooth, predictable paths (streamlines) that do not cross each other. The velocity of the fluid at any point remains constant over time. Turbulent Flow: The fluid particles move in irregular, chaotic paths, leading to eddies and swirls. The velocity of the fluid at a given point varies randomly and rapidly with time. What is terminal velocity? Write an expression for terminal velocity of a small spherical body falling through a viscous liquid. Answer: Terminal Velocity: It is the constant maximum velocity attained by an object falling through a fluid when the sum of the drag force and buoyancy force equals the downward force of gravity. Expression: $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$, where $r$ is the radius of the sphere, $\rho$ is the density of the sphere, $\sigma$ is the density of the fluid, $g$ is acceleration due to gravity, and $\eta$ is the coefficient of viscosity. State Kepler's laws of planetary motion. Answer: Kepler's First Law (Law of Orbits): All planets move in elliptical orbits with the Sun at one of the foci. Kepler's Second Law (Law of Areas): The line joining any planet to the Sun sweeps out equal areas in equal intervals of time. Kepler's Third Law (Law of Periods): The square of the orbital period ($T$) of any planet is directly proportional to the cube of the semi-major axis ($a$) of its orbit. ($T^2 \propto a^3$) Define Young's modulus. Write its unit. Answer: Young's Modulus ($Y$): It is a measure of the stiffness of an elastic material. It is defined as the ratio of tensile (or compressive) stress to longitudinal strain. $Y = \frac{\text{Tensile Stress}}{\text{Longitudinal Strain}}$. Unit: Pascal (Pa) or N/m$^2$. A body of mass $2 \text{ kg}$ is acted upon by a force of $10 \text{ N}$. What is the acceleration produced? Answer: Given: $m = 2 \text{ kg}$, $F = 10 \text{ N}$. Using Newton's second law, $F = ma$. $10 = 2 \times a$ $a = \frac{10}{2} = 5 \text{ m/s}^2$. What is meant by the range of a projectile? Write the expression for the range. Answer: Range ($R$): The horizontal distance covered by a projectile from its point of projection to the point where it hits the ground (or the same horizontal level). Expression: $R = \frac{u^2 \sin(2\theta)}{g}$, where $u$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity. Differentiate between scalar and vector quantities with examples. Answer: Scalar Quantity: A physical quantity that has only magnitude but no direction. Example: Mass, time, distance, speed, temperature. Vector Quantity: A physical quantity that has both magnitude and direction. Example: Displacement, velocity, acceleration, force, momentum. Part C: Answer any 6 questions. Each question carries 3 marks. Derive the expression for the centripetal acceleration of a body performing uniform circular motion. Answer: Consider a particle moving in a circle of radius $r$ with uniform speed $v$. Let the particle be at point A at time $t$ and at point B at time $t+\Delta t$. The change in velocity vector $\Delta \vec{v} = \vec{v_2} - \vec{v_1}$. From vector diagram (isosceles triangle formed by $\vec{v_1}$, $\vec{v_2}$, and $\Delta \vec{v}$), and considering similar triangles (position vectors $\vec{r_1}$, $\vec{r_2}$, and $\Delta \vec{r}$), we have: $\frac{|\Delta \vec{v}|}{|\vec{v}|} = \frac{|\Delta \vec{r}|}{|\vec{r}|}$ $\frac{\Delta v}{v} = \frac{\Delta r}{r}$ $\Delta v = \frac{v}{r} \Delta r$ Centripetal acceleration $a_c = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \lim_{\Delta t \to 0} \frac{v}{r} \frac{\Delta r}{\Delta t}$ Since $\lim_{\Delta t \to 0} \frac{\Delta r}{\Delta t} = v$ (speed), $a_c = \frac{v}{r} \cdot v = \frac{v^2}{r}$. In terms of angular velocity $\omega$: $v = r\omega$, so $a_c = \frac{(r\omega)^2}{r} = r\omega^2$. Prove that for a projectile, the maximum height $H = \frac{u^2 \sin^2 \theta}{2g}$. Answer: Consider the vertical motion of the projectile. Initial vertical velocity $u_y = u \sin \theta$. Final vertical velocity at maximum height $v_y = 0$. Acceleration $a_y = -g$. Using the kinematic equation: $v_y^2 = u_y^2 + 2a_y H$ $0^2 = (u \sin \theta)^2 + 2(-g)H$ $0 = u^2 \sin^2 \theta - 2gH$ $2gH = u^2 \sin^2 \theta$ $H = \frac{u^2 \sin^2 \theta}{2g}$. State and prove the law of conservation of mechanical energy for a freely falling body. Answer: Law of Conservation of Mechanical Energy: For a conservative force (like gravity), the total mechanical energy (sum of kinetic and potential energy) of a system remains constant, provided no non-conservative forces (like air resistance) do work on it. $E = K + U = \text{constant}$. Proof for a freely falling body: Consider a body of mass $m$ falling freely under gravity. Let its initial height be $h$ and it falls to the ground (height 0). At height $h$ (Point A): Initial velocity $u = 0$ (assuming dropped from rest). Kinetic Energy $K_A = \frac{1}{2}mu^2 = 0$. Potential Energy $U_A = mgh$. Total Mechanical Energy $E_A = K_A + U_A = 0 + mgh = mgh$. At height $x$ (Point B) from the ground: Let its velocity be $v$. Using $v^2 = u^2 + 2as$: $v^2 = 0^2 + 2g(h-x) = 2g(h-x)$. Kinetic Energy $K_B = \frac{1}{2}mv^2 = \frac{1}{2}m(2g(h-x)) = mg(h-x)$. Potential Energy $U_B = mgx$. Total Mechanical Energy $E_B = K_B + U_B = mg(h-x) + mgx = mgh - mgx + mgx = mgh$. At the ground (Point C, $x=0$): Let its velocity be $V$. Using $V^2 = u^2 + 2as$: $V^2 = 0^2 + 2gh = 2gh$. Kinetic Energy $K_C = \frac{1}{2}mV^2 = \frac{1}{2}m(2gh) = mgh$. Potential Energy $U_C = mg(0) = 0$. Total Mechanical Energy $E_C = K_C + U_C = mgh + 0 = mgh$. Since $E_A = E_B = E_C = mgh$, the total mechanical energy is conserved during free fall. Define escape velocity. Derive an expression for escape velocity. Answer: Escape Velocity ($v_e$): It is the minimum velocity with which a body must be projected vertically upwards from the surface of a planet so that it escapes the gravitational field of the planet and never returns. Derivation: Consider a body of mass $m$ projected from the Earth's surface (mass $M_e$, radius $R_e$). At the surface, the total energy of the body is: $E_1 = K_1 + U_1 = \frac{1}{2}mv_e^2 - \frac{GM_e m}{R_e}$. For the body to escape, its final velocity at infinite distance from Earth must be zero, and its potential energy at infinity is considered zero. $E_2 = K_2 + U_2 = 0 + 0 = 0$. By the principle of conservation of energy: $E_1 = E_2$. $\frac{1}{2}mv_e^2 - \frac{GM_e m}{R_e} = 0$ $\frac{1}{2}mv_e^2 = \frac{GM_e m}{R_e}$ $v_e^2 = \frac{2GM_e}{R_e}$ $v_e = \sqrt{\frac{2GM_e}{R_e}}$. Since $g = \frac{GM_e}{R_e^2}$, we can write $GM_e = gR_e^2$. So, $v_e = \sqrt{\frac{2gR_e^2}{R_e}} = \sqrt{2gR_e}$. Explain Bernoulli's principle. State its applications. Answer: Bernoulli's Principle: For an incompressible, non-viscous fluid in streamline flow, the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant at all points along a streamline. Mathematically: $P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$ Where $P$ is pressure, $\rho$ is fluid density, $v$ is fluid velocity, $g$ is acceleration due to gravity, and $h$ is height. Applications: Aerofoil (Lift on aircraft wings): The shape of the wing causes air to flow faster over the top surface than the bottom, leading to lower pressure above and higher pressure below, creating lift. Venturimeter: Used to measure the flow speed of an incompressible fluid. Atomizer/Sprayer: High-speed air creates low pressure, drawing liquid up to be sprayed. Bunsen burner: Gas drawn into the air stream due to low pressure. Magnus effect: Explains the curveball in sports (spinning ball creates pressure difference). Derive an expression for the potential energy of a mass $m$ placed at a height $h$ from the ground. Answer: Potential energy is the energy stored in an object due to its position or configuration. Gravitational potential energy is stored due to its position in a gravitational field. Consider a body of mass $m$ lifted vertically upwards to a height $h$ against the gravitational force. The gravitational force acting on the body is $F_g = mg$ (downwards). To lift the body, an external force $F_{ext}$ equal in magnitude to $F_g$ must be applied upwards, i.e., $F_{ext} = mg$. Work done by this external force in lifting the body to height $h$ is: $W = F_{ext} \times \text{displacement} = mg \times h = mgh$. This work done against gravity is stored in the body as its gravitational potential energy ($U$). Therefore, $U = mgh$. A force $\vec{F} = 3\hat{i} + 4\hat{j} \text{ N}$ acts on a particle. The displacement of the particle is $\vec{d} = 3\hat{i} + 4\hat{j} \text{ m}$. Calculate the work done by the force. Answer: Work done $W = \vec{F} \cdot \vec{d}$. $W = (3\hat{i} + 4\hat{j}) \cdot (3\hat{i} + 4\hat{j})$ $W = (3)(3) + (4)(4)$ $W = 9 + 16$ $W = 25 \text{ Joules}$. State parallel axis theorem. Use it to find the moment of inertia of a ring about a tangent in its plane. Answer: Parallel Axis Theorem: It states that the moment of inertia of a rigid body about any axis is equal to its moment of inertia about a parallel axis passing through its center of mass plus the product of the mass of the body and the square of the perpendicular distance between the two axes. $I = I_{CM} + Md^2$. Moment of inertia of a ring about a tangent in its plane: For a ring of mass $M$ and radius $R$: 1. Moment of inertia about an axis passing through its center and perpendicular to its plane is $I_Z = MR^2$. 2. By perpendicular axis theorem, $I_Z = I_X + I_Y$. For a ring, due to symmetry, $I_X = I_Y$. So, $MR^2 = 2I_X \implies I_X = \frac{1}{2}MR^2$. This $I_X$ is the moment of inertia about a diameter (which passes through CM and is in the plane of the ring). 3. Now, we want to find the moment of inertia about a tangent in its plane. This tangent is parallel to a diameter, and the distance between them is $d = R$. Applying the parallel axis theorem: $I_{\text{tangent}} = I_X + Md^2$ $I_{\text{tangent}} = \frac{1}{2}MR^2 + M(R)^2$ $I_{\text{tangent}} = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$. Part D: Answer any 2 questions. Each question carries 4 marks. a) What is banking of roads? Why is it necessary? b) Derive an expression for the maximum safe speed for a car on a banked road without relying on friction. Answer: a) Banking of Roads: It is the phenomenon of raising the outer edge of a curved road above its inner edge. Necessity: When a vehicle takes a turn on a flat road, the necessary centripetal force is provided by the force of friction between the tires and the road. If the speed of the vehicle is high or the radius of the turn is small, the required centripetal force may exceed the maximum static friction, causing the vehicle to skid. Banking the road helps to provide the necessary centripetal force partly by the normal force component, reducing the reliance on friction and allowing for safer turns at higher speeds. b) Expression for maximum safe speed on a banked road without friction: Consider a vehicle of mass $m$ moving on a banked road with angle $\theta$. The forces acting on the vehicle are: 1. Weight $mg$ acting vertically downwards. 2. Normal force $N$ acting perpendicular to the banked surface. Since there is no friction, the only force that can provide the centripetal force is the horizontal component of the normal force. Resolve $N$ into horizontal and vertical components: Vertical component: $N \cos \theta$ (balances weight) Horizontal component: $N \sin \theta$ (provides centripetal force) From vertical equilibrium: $N \cos \theta = mg \quad \ldots (1)$ From centripetal force requirement: $N \sin \theta = \frac{mv^2}{r} \quad \ldots (2)$ Divide equation (2) by equation (1): $\frac{N \sin \theta}{N \cos \theta} = \frac{mv^2/r}{mg}$ $\tan \theta = \frac{v^2}{rg}$ $v^2 = rg \tan \theta$ $v = \sqrt{rg \tan \theta}$. This is the optimum speed for a banked road, where no friction is needed to negotiate the turn. a) State Newton's Law of Gravitation. b) Define gravitational potential energy. Derive an expression for the gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of Earth (mass $M$). Answer: a) Newton's Law of Gravitation: Every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Mathematically: $F = G \frac{m_1 m_2}{r^2}$, where $G$ is the universal gravitational constant. b) Gravitational Potential Energy: It is the amount of work done in bringing a body from infinity to a given point in the gravitational field without acceleration. The potential energy at infinity is taken as zero. Derivation: Consider a body of mass $m$ being brought from infinity ($\infty$) to a point at a distance $r$ from the center of Earth (mass $M$). The gravitational force acting on the body at a distance $x$ from the center of Earth is $F(x) = \frac{GMm}{x^2}$. To bring the body inward, an external force equal and opposite to $F(x)$ must be applied. The work done by the external force in moving the body by a small distance $dx$ is $dW = F(x) dx = \frac{GMm}{x^2} dx$. The total work done in bringing the body from infinity to distance $r$ is: $W = \int_{\infty}^{r} \frac{GMm}{x^2} dx$ $W = GMm \int_{\infty}^{r} x^{-2} dx$ $W = GMm \left[ \frac{x^{-1}}{-1} \right]_{\infty}^{r}$ $W = GMm \left[ -\frac{1}{x} \right]_{\infty}^{r}$ $W = GMm \left( -\frac{1}{r} - (-\frac{1}{\infty}) \right)$ $W = GMm \left( -\frac{1}{r} - 0 \right)$ $W = -\frac{GMm}{r}$. This work done is stored as the gravitational potential energy ($U$) of the body at distance $r$. Therefore, $U = -\frac{GMm}{r}$. The negative sign indicates that the force is attractive and that potential energy decreases as the objects get closer. a) State and prove the principle of conservation of energy for a simple harmonic motion. b) What is the period of a simple pendulum of length $1 \text{ m}$? (Take $g = 9.8 \text{ m/s}^2$) Answer: a) Principle of Conservation of Energy for SHM: For a simple harmonic oscillator, the total mechanical energy (sum of kinetic and potential energy) remains constant throughout its motion, provided there are no dissipative forces like friction or air resistance. Proof: Consider a particle of mass $m$ executing SHM. The displacement of the particle is given by $x(t) = A \cos(\omega t + \phi)$. The velocity of the particle is $v(t) = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi)$. Kinetic Energy ($K$): $K = \frac{1}{2}mv^2 = \frac{1}{2}m(-A\omega \sin(\omega t + \phi))^2 = \frac{1}{2}mA^2\omega^2 \sin^2(\omega t + \phi)$. Potential Energy ($U$): For SHM, the restoring force is $F = -kx$. Potential energy $U = \int F_{ext} dx = \int kx dx = \frac{1}{2}kx^2$. Since for SHM, $k = m\omega^2$, we have: $U = \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2 (A \cos(\omega t + \phi))^2 = \frac{1}{2}mA^2\omega^2 \cos^2(\omega t + \phi)$. Total Mechanical Energy ($E$): $E = K + U$ $E = \frac{1}{2}mA^2\omega^2 \sin^2(\omega t + \phi) + \frac{1}{2}mA^2\omega^2 \cos^2(\omega t + \phi)$ $E = \frac{1}{2}mA^2\omega^2 [\sin^2(\omega t + \phi) + \cos^2(\omega t + \phi)]$ As $\sin^2\theta + \cos^2\theta = 1$: $E = \frac{1}{2}mA^2\omega^2$. Since $m$, $A$, and $\omega$ are constants for a given SHM, the total mechanical energy $E$ is constant and independent of time. This proves the conservation of energy for SHM. b) Period of a simple pendulum: The formula for the period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$. Given: $L = 1 \text{ m}$, $g = 9.8 \text{ m/s}^2$. $T = 2\pi \sqrt{\frac{1}{9.8}}$ $T \approx 2 \times 3.14159 \times \sqrt{0.10204}$ $T \approx 6.28318 \times 0.3194$ $T \approx 2.00 \text{ seconds}$. The period of the simple pendulum is approximately $2 \text{ seconds}$.